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3 years ago

5

16x+9=9y-2x Solve for y

1 answer:

Novay_Z [31]3 years ago

8 0

Y=2x+1... I hope this helps love! :)

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I need the answer!!!

Maslowich

Answer:

The first one is 12

second one is 1

third one is 23

last one 5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Cos(2theta)=11 sin(theta)-5

lubasha [3.4K]

$\cos2\theta=11\sin\theta-5$
$1-2\sin^2\theta=11\sin\theta-5$
$2\sin^2\theta+11\sin\theta-6=0$
$(2\sin\theta-1)(\sin\theta+6)=0$
$\implies\begin{cases}2\sin\theta-1=0\\\sin\theta+6=0\end{cases}$

The second equation gives $\sin\theta=-6$ , but $-1\le\sin\varphi\le1$ for any real $\varphi$ , so we can ignore this equation.

This leaves us with

$2\sin\theta-1=0\implies \sin\theta=\dfrac12\implies\theta=\dfrac\pi6+2\pi k,\theta=\dfrac{5\pi}6+2\pi k$

where $k$ is any integer.

3 0

3 years ago

If the diagonals of a quadrilateral do not bisect each other then the quadrilateral could be a

Svet_ta [14]

Trapezoid. The diagonals of all parallelograms bisect each other. Trapezoid are not parallelograms.

8 0

3 years ago

Consider a Triangle ABC like the one below. Suppose that C = 98, A = 74, and b = 11 (figure is not drawn to scale.) solve the tr

Degger [83]

Answer:

$A=73.8\°$

$B=8.2\°$

$c=76.3\ units$

Step-by-step explanation:

step 1

Find the measure of side c

Applying the law of cosines

$c^{2}= a^{2}+b^{2}-2(a)(b)cos(C)$

substitute the given values

$c^{2}= 74^{2}+11^{2}-2(74)(11)cos(98\°)$

$c^{2}=5,823.5738$

$c=76.3\ units$

step 2

Find the measure of angle A

Applying the law of sine

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

substitute the given values

$\frac{74}{sin(A)}=\frac{76.3}{sin(98\°)}$

$sin(A)=(74)sin(98\°)/76.3$

$A=arcsin((74)sin(98\°)/76.3)$

$A=73.8\°$

step 3

Find the measure of angle B

we know that

The sum of the internal angles of a triangle must be equal to 180 degrees

so

$A+B+C=180\°$

substitute the given values

$73.8\°+B+98\°=180\°$

$171.8\°+B=180\°$

$B=180\°-171.8\°=8.2\°$

5 0

3 years ago

The Wall Street Journal reports that the rate on three-year Treasury securities is 7.00 percent, and the six-year Treasury rate

Art [367]

Maturity risk for 3 year Treasury security is 0.1%

<u>Explanation:</u>

Given -

Rate of 3 year Treasury security, R3 = 7%

Rate of 6 year Treasury security, R6 = 6.2%

Rate of inflation, Ri = 2.25% next year, 2.5% on 2nd year, 2.5% on 3rd year

Real interest rate, Rint = 4.4%

Maturity risk on 3 year Treasury security, x = ?

Maturity risk for 3 year Treasury security:

R3 = Ri on 3rd year + Rint + x

7% = 2.5% + 4.4% + x

x = 0.1%

Therefore, Maturity risk for 3 year Treasury security is 0.1%

8 0

4 years ago